On the Independence Number of the Erd˝Os-Rényi And

ON THE INDEPENDENCE NUMBER OF THE ERDOS-R˝ ENYI´ AND PROJECTIVE NORM GRAPHS AND A RELATED HYPERGRAPH

DHRUV MUBAYI∗ AND JASON WILLIFORD

Abstract. The Erd˝os-Rényi and Projective Norm graphs are algebraically defined graphs that have proved useful in supplying constructions in extremal graph theory and Ramsey theory. Their eigenvalues have been computed and this yields an upper bound on their independence number. Here we show that in many cases, this upper bound is sharp in order of magnitude. Our result for the Erd˝os-Rényi graph has the following reformulation: the maximum size of a family of mutually non-orthogonal lines in a vector space of dimension three over the finite field of order q is of order q3/2. We also prove that every subset of vertices of size greater than q2/2 + q3/2 + O(q) in the Erd˝os-Rényi graph contains a triangle. This shows that an old construction of Parsons is asymptotically sharp. Several related results and open problems are provided.

1. Introduction The independence number α(G) of a graph or hypergraph G is the maximum size of a subset of vertices of G that contains no edge. The eigenvalues of the adjacency matrix of a graph provide an upper bound on its independence number. The aim of this paper is to examine the tightness of these bounds when applied to several well- known families of graphs. The particular graphs considered are the Erd˝os-R´enyi and Projective Norm graphs, and various subgraphs of these. We also consider a related hypergraph obtained from the Erd˝os-R´enyi graph that has recently proved useful in extremal hypergraph theory. Given a graph G and any set I ⊂ V (G) let e(I,I) be the number of ordered pairs of vertices in I which are adjacent. The aforementioned eigenvalue bound, found, for example, in [4], gives the following. Theorem 1. Let λ be the second largest eigenvalue in absolute value of the adjacency matrix of a d-regular graph G (possibly with loops) with n vertices. Then d 2 |e(I,I) − n |I| | ≤ λ|I|. In particular, if I contains no edges, except for possibly loops, then |I| ≤ (λ + 1)n/d.

1.1. The Projective Norm Graphs. Let t > 1 be a positive integer, Fq be the ∗ finite field of order q and Fq be the multiplicative group of Fq. The norm from 1+q+···+qt−2 Fqt−1 to Fq is the function N : Fqt−1 → Fq defined by N(X) = X .

∗Research supported in part by National Science Foundation grant DMS-0400812, and an Alfred P. Sloan Research Fellowship. 1 2 DHRUV MUBAYI∗ AND JASON WILLIFORD

Definition. Fix an integer t ≥ 2 and a prime power q. The Projective Norm graph ∗ G = Gq,t has vertex set V (G) = Fqt−1 × Fq with two vertices (A, a), (B, b) ∈ V (G) adjacent when N(A + B) = ab. For a graph G let ex(n, G) deonte the largest number of edges of any graph on n vertices which has no copy of G as a subgraph. Let Rk(G1,...,Gk) denote the smallest integer n such that any edge coloring in k colors of Kn must contain an i colored copy of Gi for some 1 ≤ i ≤ k, and let Rk(G) = Rk(G, . . . , G). The graphs Gq,t were first constructed by Alon, Rónyai and Szabóto give improved lower bounds for the Turánfunction ex(n, Kt,s) and the Ramsey numbers Rk(Kt,s) where t ≥ 2 and s ≥ (t − 1)! + 1 (see [3]). In [22] T. Szabófound the eigenvalues of Gq,t to be ±q(t−1)/2, ±1, 0 and qt−1 −1. This was done independently by Alon and Rödlin [2], who used these eigenvalues to provide tight bounds for Rk(Kt,s,...,Kt,s,Km) where again t ≥ 2 and s ≥ (t − 1)! + 1 (see [3]). t t−1 t−1 The projective norm graph Gq,t has n = q − q vertices, and degree q − 1. Using the results of [2, 22] for its eigenvalues and Theorem 1, we obtain for fixed t ≥ 2 (qt − qt−1)(q(t−1)/2 + 1) (1.1) α(G ) ≤ = (1 + o(1))q(t+1)/2 as q → ∞. q,t qt−1 − 1 Our first result shows that (1.1) gives the correct order of magnitude for all odd t. Theorem 2. Let t > 1 be an odd integer and q an odd prime power. Then q(t+1)/2 − q(t−1)/2 α(G ) ≥ . q,t 2 Thus as q → ∞, (t+1)/2 (t+1)/2 (1/2 + o(1))q < α(Gq,t) < (1 + o(1))q . The problem of determining whether (1.1) is sharp for even t seems to be more difficult. Open Problem 1. Find a construction of an independent set of size Cq(t+1)/2, C a constant, for even values of t > 2 or q a power of 2. 1.2. The Erd˝os-Rényi Graph. We will pay special attention to the Projective Norm graphs in the case t = 2. The graph Gq,2 is a large induced subgraph of a well-known graph called the Erd˝os-Rényi graph which we denote ERq. Since the graph ERq is of independent interest, we begin by explaining its construction and connection to Gq,2. Let q be a prime power and V be a 3-dimensional vector space over a finite field Fq. The projective geometry PG(2, q) is the triple (P, L, I) where P is the set of all 1-dimensional subspaces of V which we call points, L is the set of all 2-dimensional subspaces which we call lines, and the incidence relation I is containment. Points of PG(2, q) will be represented by left-normalized vectors, vectors whose left-most non-zero entry is 1, and which span the 1-dimensional space in question. Similarly lines will be represented by left-normalized vectors which span the orthogonal com- plement of the corresponding 2-dimensional subspace of V . We use round brackets for points and square brackets for lines to avoid confusion. Therefore a point (x0, x1, x2) is on a line [y0, y1, y2] if and only if x0y0 + x1y1 + x2y2 = 0. A polarity φ of a projective plane PG(2, q) is a bijective map from P ∪ L to itself that maps INDEPENDENCE NUMBER OF PROJECTIVE NORM AND RELATED GRAPHS 3 points to lines and lines to points and reverses incidence (meaning p ∈ l if and only if φ(l) ∈ φ(p)) and with the property that φ2 is the identity map on π. A point p is called absolute with respect to the polarity φ if p ∈ φ(p). The polarity graph of PG(2, q) with respect to a polarity φ is the simple graph (V,E) with vertex set equal to P such that for x, y ∈ P with x 6= y, {x, y} ∈ E if and only if x ∈ φ(y). (It should be noted that polarity graphs can alternately be defined with loops at each of the absolute points). The projective plane PG(2, q) is known to have the orthogonal polarity φo : (x , x , x ) 7→ [x , x , x ] and if q is a perfect square the unitary polarity φ : 0 1 2 √0 1√ 2 √ u q q q (x0, x1, x2) 7→ [x0 , x1 , x2 ]. Any other polarity of PG(2, q) is projectively equivalent to one of these forms (see [18]).

Definition. For q a prime power, the Erd˝os-Rényi graph ERq is the orthogonal polarity graph of PG(2, q). Formally, the vertex set of ERq is the set of points of PG(2, q) with two distinct vertices (x0, x1, x2) and (y0, y1, y2) adjacent if and only o if x0y0 + x1y1 + x2y2 = 0. We define ERq to be the graph ERq with loops attached to the absolute points.

The graph ERq was introduced in this form by Erd˝osand Rényi in [8] to give constructive examples of graphs with small maximum degree, relatively few edges and diameter 2. The graph ERq plays a notable role in extremal graph theory. Recall that if G is a graph, ex(n, G) denotes the greatest number of edges a graph on n vertices can have without containing G as a subgraph. Any graph on n vertices with ex(n, G) edges and which has no copy of G as a subgraph is called extremal. The asymptotic behavior of ex(n, G) is well understood if G is not bipartite; however, very little is known when G is bipartite (see [5] and [10] and the references therein). Of particular interest is the behavior of ex(n, C2k) where C2k denotes a cycle of length 2k. In [9] Erd˝os,Rényi and Sósproved that ex(n, C4) is 1 3/2 asymptotic with 2 n using the graphs ERq for constructive lower bounds (this was done independently by Brown in [6]). Füredi later demonstrated in [12], [13] and [14] that the graphs ERq are extremal. Such exact results are relatively rare in extremal graph theory. The graph ERq has also been used to solve a similar problem for hypergraphs (see [19]). These hypergraphs will be dealt with in the last section. Many of our results require algebraic manipulations for which ERq is not suited. This leads us to the following ∗ Definition. For q an odd prime power, ERq is the graph whose vertex set is V (ERq) with two vertices (x0, x1, x2) and (y0, y1, y2) adjacent if and only if x0y2 − x1y1 + x2y0 = 0. Several of our calculations will be aided by the following fact:

∗ Proposition 3. The graph ERq is isomorphic to ERq.

The connection between the projective norm graph Gq,2 and ERq is given by the following Proposition.

Proposition 4. The graph Gq,2 is isomorphic to an induced subgraph of ERq. 2 2 ERq has q +q +1 vertices while Gq,2 has q −q. As Gq,2 is an induced subgraph of ERq missing 2q +1 of the vertices of ERq, their independence numbers are equal up to a linear (in q) error term. The question of determining the independence 4 DHRUV MUBAYI∗ AND JASON WILLIFORD number of ERq can be phrased as a simple question about vector spaces which seems independently interesting: What is the maximum number of mutually non-orthogonal 1-dimensional subspaces of a 3-dimensional vector space over Fq? o √ The eigenvalues of ERq are well known to be ± q, q + 1 (see [1]). It is clear o that the maximum size of a set of vertices of ERq with no edges except for possibly loops is equal to α(ER ). From Theorem 1, and noting that there are q + 1 loops q √ o 3/2 q+1 in ERq, we obtain α(ERq) ≤ q + q + q+1 . Since one of our results will be close to exact, we will introduce a bound due to Hoffman (see [17]) which is sharper than Theorem 1 with respect to lower order terms. However, this bound only bounds the size of the largest independent set which has no loops, so we must add the number of loops to the result. This bound states that if λ is the smallest eigenvalue of the adjacency matrix of a d-regular n-vertex graph G, then −nλ α(G) ≤ d − λ In the case of ERq we must add q + 1 to the bound, which after simplification gives us: 3/2 √ α(ERq) ≤ q + q + 1 Godsil and Newman generalize this bound if a restricted number of absolute points are allowed, see [15] for details. We will establish that the Hoffman bound gives the correct magnitude for α(ERq). Theorem 5. Let p be a prime, n a positive integer and q = pn. Then  q3/2+q+2  2 for p odd, n even  3/2  120√q for p odd, n odd 73 73 α(ERq) ≥ 3/2  q√ for p = 2, n odd  2 2 √  q3/2 − q + q for p = 2, n even

3/2 In all cases, α(ER ) ≥ 120√q > .19239 q3/2. q 73 73

For even powers of 2 we can state exactly the size of the largest independent set which contains no absolute points. Theorem 6. Let q be an even power of 2. Then the size of the largest independent 3/2 √ set of ERq containing no absolute vertex is q − q + q. Given the level of precision in the previous result, it is natural to ask the following:

Open Problem 2. Construct an independent set I in ERq for all q which are not even powers of two such that |I| = q3/2 + O(q), or prove that no such set exists.

1.3. The Polarity Graph Uq. We next consider a graph closely related to ERq whose independence number can be found exactly for all q which are even powers of primes.

Deﬁnition. Let q be a square prime power. The unitary polarity graph Uq of PG(2, q) is the graph with vertex set V (ER ) with two vertices (x , x , x ) and √ q√ √ 0 1 2 q q q (y0, y1, y2) adjacent if and only if x0y0 + x1y1 + x2y2 = 0. INDEPENDENCE NUMBER OF PROJECTIVE NORM AND RELATED GRAPHS 5

As stated earlier, the graph Uq is the only other polarity graph which PG(2, q) admits. Uq is similar to ERq in that it has no 4-cycles and has diameter 2 (see [23] for a similar argument for ERq). However, it has fewer edges than ERq and so does not play the same role in extremal graph theory. Consequently, Uq is not as well known as ERq. As in ERq, there are absolute points. The graph formed − by deleting the absolute vertices of Uq will be denoted by Uq . Uq has precisely 3/2 q + 1 such absolute points, and these form an independent set in the graph Uq. We will show that this is the unique maximum independent set of Uq. This is in stark contrast to the graph ERq for which exact results remain elusive. To get the level of precision necessary in the upper bound of α(Uq) we will use a combinatorial bound as opposed to eigenvalue methods.

3/2 Theorem 7. Let q be a prime power. Then α(Uq) = q + 1, with the set of absolute points being the unique independent set of size q3/2 + 1.

It is interesting to note that the largest independent set in Uq consists of all the absolute points of Uq, while the best construction for ERq contains no absolute points. With the independence number of Uq resolved, it would be interesting to − see what is the magnitude of the largest independent set I of Uq (which contains no absolute points). Using the Hoﬀman bound and the fact that Uq has the same eigenvalues as ERq, we have the inequality

− 3/2 √ α(Uq ) ≤ q − q + q. From direct computation, this bound is not tight. For q = 4, 9 we have upper − − bounds of 6 and 21 while α(U4 ) = 4 and α(U9 ) = 19. A better upper bound and new constructive techniques are required to see if the size of I can also be found exactly as α(Uq).

1.4. The Erd˝os-Rényi Hypergraph of Triangles. In [19] Lazebnik, and Ver- straëte(using an idea of Lovász) construct a series of hypergraphs Hq of girth 5. These hypergraphs are used to determine the asymptotics of the Turánnumber T3(n, 8, 4), defined as the maximum number of edges in a 3-graph on n vertices in which no set of 8 vertices spans more than 4 edges.

Deﬁnition. Let q be an odd prime power. Hq is the 3-graph whose vertex set is the set of nonabsolute points of V (ERq) and edge set is the set of triangles in ERq.

It is apparent from the construction of Hq that α(Hq) is the order of the largest triangle-free induced subgraph of ERq which contains no absolute points. In [21] q+1 Parsons constructs such a subgraph which has 2 vertices if q ≡ 3 (mod 4) and q 2 vertices if q ≡ 1 (mod 4) (Parsons’ aim was to study the automorphism groups of these subgraphs). We will show that Parsons’ construction is asymptotically tight using eigenvalue techniques.

2 3/2 Theorem 8. Let q be an odd prime power. Then α(Hq) ≤ q /2 + q + O(q). Thus in particular, 2 α(Hq) = (1/2 + o(1))q . Having shown that Parson’s graphs are asymptotically tight, we conjecture the following. 6 DHRUV MUBAYI∗ AND JASON WILLIFORD

q Conjecture 1. Let q be an odd prime power. Then α(Hq) = 2 if q ≡ 1 (mod 4) q+1 and α(Hq) = 2 if q ≡ 3 (mod 4). The only constructions known of size roughly q2/2 are for odd q.

Open Problem 3. Let q be a power of 2. Find an induced subgraph of ERq which is triangle free and has at least q2/2 + O(q3/2) vertices. The proofs of all stated results are in the following sections, labelled according to the graphs they pertain to.

2. The Projective Norm Graph Gq,t Proof of Theorem 2. Recall that t > 1 is odd, q is an odd prime power, and we q(t+1)/2−q(t−1)/2 are to show that α(Gq,t) ≥ 2 . Let µ be a primitive element of Fqt−1 and set (t−1)/2 E = {µs(q +1) : s ∈ [1, q(t−1)/2 − 1]} ∪ {0}.

Since t − 1 is even, there is a subﬁeld Fq(t−1)/2 of Fqt−1 which consists of those q(t−1)/2 x ∈ Fqt−1 which satisfy x = x. It is easy to verify this property for elements of E and therefore E is closed under addition. One also notes that every element s(q(t−1)/2+1)/2 2 x of E is a square in Fqt−1 since x is 0 or x = (µ ) for some s ∈ [1, q(t−1)/2 − 1]}. Let S = {µx : x ∈ E} and let T be the set of all nonzero squares in Fq. We will show that I = S × T is an independent set in Gq,t. First observe that S is closed under addition because E is, and every element of S is a nonsquare as it is the product of µ and a square. Also, T is closed under multiplication as the product of two nonzero squares in Fq is a nonzero square in Fq. Let A, B ∈ S and a, b ∈ T . Then A + B is a nonsquare in Fqt−1 and therefore N(A+B) is a nonsquare in Fq. As ab is a square in Fq, we must have N(A+B) 6= ab and therefore I is an independent set of size |S||T | = q(t−1)/2(q −1)/2 = (q(t+1)/2 − q(t−1)/2)/2.

3. The Erdos-R˝ enyi´ Graph ERq In this section we completely prove Theorem 5. We begin be proving Propositions 3 and 4, both of which are needed in the proof of Theorem 5. ∗ Proof of Proposition 3. We must show that ERq is isomorphic to ERq, where ∗ ERq is the graph with the same vertex set as ERq but with two vertices x and y connected when x0y2 − x1y1 + x2y0 = 0. We need a change of basis of V which transforms the form x0y0 +x1y1 +x2y2 = 0 T into x0y2−x1y1+x2y0 = 0. This requires a 3 by 3 matrix C where CMC = λI, λ ∈ 0 0 1 Fq, λ 6= 0, where M = 0 −1 0. We explicitly give matrices C satisfying these 1 0 0 properties. 1 1 1 If q is a power of 2, we use the matrix C2 = 0 1 1. If q is odd, let a, b, c, d, i 1 1 0 be such that a2 = −2, b2 = 2, c2 + d2 = −1, i2 = −1 (when they exist). We use INDEPENDENCE NUMBER OF PROJECTIVE NORM AND RELATED GRAPHS 7 the following change of variables for q ≡ 1 (mod 4) and q ≡ 3, 7 (mod 8) which we label C1,C3,C7 respectively:  (1+i) (1−i)   a a   1 1  2 0 2 2 a 2 b 0 b −d d C1 =  0 i 0  C3 = −1 −1 −1 C7 =  b c b  (−1+i) (1+i) −a 0 a c d −c 2 0 − 2 2 2 b b Proof of Proposition 4. We must show that Gq,2 is isomorphic to a sub- ∗ graph of ERq. Recall that Gq,2 has vertex set V (G) = Fq × Fq with two vertices (A, a), (B, b) ∈ V adjacent when N(A + B) = ab. As the norm from Fq to itself is trivial, the norm may be omitted and we have that two vertices are connected ∗ when A + B = ab. Having established that ERq and ERq are isomorphic, now ∗ take the set W of vertices of ERq of the form (1, x1, x2) where x1 6= 0. Let Hq be ∗ the subgraph of ERq induced by W . We form a map φ : Hq → Gq,2 deﬁned by φ : (1, x1, x2) 7→ (x2, x1). This map is an isomorphism as two vertices x, y of Hq are adjacent if and only if x2 − x1y1 + y2 = 0, which is equivalent to x2 + y2 = x1y1. Proof of Theorem 5. Let p be prime, n > 0, and q = pn. We prove all cases here except when p = 2 and n is even. For this case we obtain sharper results, which we postpone to the next subsection. By virtue of Proposition 3 we will work with ∗ ERq instead of ERq. Let µ be a primitive element of Fq. √ √ n ( q+1)k h q−3 io Case (i): p > 2 and n is even. Let R = µ : k ∈ 0, 2 ∪ {0}. Then √ R is isomorphic to a subset of F q which has the property that for x, y ∈ R, x = −y √ implies that x = y = 0. This follows as −1 = µ(q−1)/2. If x 6= 0 then x = µ( q+1)k √ √ √ √ h q−3 i ( q+1)k (q−1)/2 ( q+1)(k+( q−1)/2) where k ∈ 0, 2 , and −x = µ µ = µ 6∈ R since √ √ h q−3 i (k + ( q − 1)/2) ∈/ 0, 2 . We claim that 2 I = {(1, t, t /2 − µr/2) : t ∈ Fq, r ∈ R} ∪ {(0, 0, 1)}

q3/2+q+2 is an independent set of size 2 . 2 2 By way of contradiction, if (1, t, t /2 − µr1/2), (1, s, s /2 − µr2/2) ∈ I are two 2 2 adjacent vertices then st = t /2 − µr1/2 + s /2 − µr2/2. This yields µ(r1 + r2) = t2 −2st+s2 = (t−s)2. As the right hand side of the equation is a square, equality is possible only if r1 = −r2 and s = t. However, the only element of R whose additive inverse is also in R is 0. Therefore, r1 = r2 implying that these two vertices are not distinct, a contradiction.

Case (ii): p > 2 and n is odd. Write x ∈ Fq in the form n−1 X i x = xiµ i=0 where xi ∈ Fp. Let A be the set of all integers in the interval [ dp/6e, bp/2c ] and B be the set of all integers in the interval [0, bpp/3c]. We let our sets S and T be n−1 the following with m = 2 :

S = {x : xn−1 ∈ A} 8 DHRUV MUBAYI∗ AND JASON WILLIFORD

T = {y : yi = 0 if i > m and ym ∈ B} p+1 For w, x ∈ S we have (w + x)n−1 ∈ d 3 e, p − 1 . For y, z ∈ T we have

m !  m  n−1 i X i X j X X i yz = yiµ  zjµ  = yi−jzjµ i=0 j=0 i=0 j=0 p−1 This gives us (yz)n−1 = (ym)(zm) ∈ 0, b 3 c . Then I = {(1, t, s): s ∈ S and t ∈ T } is an independent set with: r 3/2 1 p p p q 1 − p 3 |I| = − + 1 + 1 √ > √ q 2 2 6 3 p p 3 3 1 1− p 3 Since the constants are bounded below by √ q 2 , it is clear that there is a 3 3 prime which yields the worst constant. In [23] a search using the software package Magma was done to determine that the worst case is when p = 73 which gives the constant 120√ . 73 73

n−1 Case (iii): p = 2 and n is odd. Let m = 2 and consider the following sets:

S = {x : xn−1 = 0}

T = {y : yi = 0 if i > m and ym = 1} 3/2 Then I = {(1, t, s): s ∈ S and t ∈ T } is an independent set of size q√ . The 2 2 remaining case where n is even and p = 2 will be veriﬁed in the following subsection. The case where n and p are both odd yields the smallest constants. All cases 3/2 considered, this veriﬁes that α(ER ) ≥ 120√q . q 73 73

3.1. Even powers of 2. We now prove that when q = 2n, n even, α(ER ) ≥ √ q q3/2 − q + q. Our construction uses a special Denniston maximal arc, for more information on maximal arcs see [7],[11], and [20]. The coordinatization of the arc will be similar to that of Mathon’s (see [20]). The following lemma is based on a √ result in [7]. We pay special attention to the subfield F q. Lemma 9. Let q be an even power of 2 and x2 + x + s be an irreducible polynomial 2 over Fq. Then the image of x + x + s is a coset of an additive subgroup of index √ 2 of Fq which is disjoint from the subfield F q. Proof. As noted in [7], for fields of characteristic two, the polynomial x2 + x has 2 2 2 the property that for x, y ∈ Fq x + x + y + y = (x + y) + (x + y) and is therefore an endomorphism of the additive group of Fq with kernel {0, 1}. The image of a polynomial of the form x2 + x + c where c is a constant is therefore a subgroup of index 2 or the coset of a subgroup of index 2, depending on whether or not c is in the image of x2 + x. Since every polynomial of the form x2 + x + c with √ 2 c ∈ F q splits in Fq, the equation x + x + c = 0 has a solution in Fq, therefore, 2 2 x + x = c has a solution in Fq. This implies that the image of x + x contains the √ 2 entire subfield F q. However, the image of x + x does not contain s due to the fact that x2 + x + s is irreducible (since any solution to x2 + x = s is a solution to INDEPENDENCE NUMBER OF PROJECTIVE NORM AND RELATED GRAPHS 9 x2 + x + s = 0); therefore, the image of x2 + x + s is a coset of the image of x2 + x. 2 √ It follows that the image of x + x + s is disjoint from F q.

2 Proof of Theorem 6. Let x + x + s be an irreducible polynomial over Fq. Let I √ be the set of all points (x0, x1, x2) ∈ ERq for which there exists λ ∈ F q such that x2 + x x + sx2 + λx2 = 0. Then by a result of Denniston [7] the set I is a maximal 2 2 0 √ 0 1 √ arc of order q. It can be argued combinatorially that |I| = q3/2 −q+ q using that I is a maximal arc, but here we count the size of the set directly. If λ = 0, we must 2 have x0 = 0 (if x0 = 1 we would have a solution to x2 + x2 + s = 0, a contradiction) and hence x2 = 0. This implies x1 = 1, and we have only one solution. In the case where λ 6= 0, we then condition on whether x0 = 0 or 1. If x0 = 0, then 2 x1 = 1 (as x1 = 0 implies x2 = 0 which is impossible). We then have x2 = λ which √ 2 gives us q − 1 solutions as every choice of a nonzero λ uniquely determines x2 and hence x2 (as squaring is a field automorphism over fields of characteristic two). Lastly if λ 6= 0 and x = 1, then x2 = λ−1(x2 + x + s) and any choice of λ 6= 0 0 1 √ 2 2 and x uniquely determines x , yielding q( q − 1) solutions. Altogether we have √2 √ 1 √ √ 1 + q − 1 + q( q − 1) = q3/2 − q + q solutions, therefore, |I| = q3/2 − q + q. ∗ We claim that this particular arc forms an independent set in ERq . Assume two √ points (x0, x1, x2) and (y0, y1, y2) are connected. Then for some λ1, λ2 ∈ F q the three equations below are satisfied: 2 2 2 x2 + x2x0 + sx0 + λ1x1 = 0 2 2 2 y2 + y2y0 + sy0 + λ2y1 = 0 x2y0 + x0y2 = x1y1 2 2 If λ1 = 0 or x1 = 0, then we have that x2 +x2x0 +sx0 = 0 which forces x0 = x2 = 0 as otherwise x2 is a root of the polynomial x2 + x + s which is impossible. We have x0 an immediate contradiction if x1 = 0, as (0, 0, 0) is not a vertex of ERq. If λ1 = 0 2 2 then we must have x1 = 1 and therefore y1 = 0. Then we obtain y2 +y2y0 +sy0 = 0 which forces y0 = y2 = 0, which implies (y1, y2, y3) = (0, 0, 0), a contradiction. If λ1, λ2, x1, y1 6= 0, then we may rewrite the above equations to obtain:

1 2 2 2 (x2 + x2x0 + sx0) = x1 λ1

1 2 2 2 (y2 + y2y0 + sy0) = y1 λ2 x0y2 + x2y0 = x1y1 Squaring the third equation and substituting we get:

2 1 2 2 2 2 (x2y0 + x0y2) = (x2 + x2x0 + sx0)(y2 + y2y0 + sy0) λ1λ2 The quantity x2y0 + x0y2 6= 0 since x1, y1 6= 0, therefore, we obtain: 2 2 2 2 (x2 + x2x0 + sx0)(y2 + y2y0 + sy0) 2 = λ1λ2 (x2y0 + x0y2) 2 If x0 = 0, then x1 = y0 = 1 and we have y2 + y2 + s = λ1λ2, which is impossible by Lemma 9. Similarly, y0 = 0 also leads to a contradiction. The last case to consider is x0 = y0 = 1, which yields the equation: 2 2 (x2 + x2 + s)(y2 + y2 + s) √ 2 = λ1λ2 ∈ F q (x2 + y2) 10 DHRUV MUBAYI∗ AND JASON WILLIFORD

Noting that y2 6= x2, we substitute y2 = 1/w + x2. After expanding we have the equation:

2 2 2 2 2 √ (x2 + x2 + s) w + (x2 + x2 + s)w + x2 + x2 + s ∈ F q which is equivalent to:

2 2 2 √ ((x2 + x2 + s)w + x2) + ((x2 + x2 + s)w + x2) + s ∈ F q This is impossible by Lemma 9. All cases accounted for, we have that no two points 3/2 √ of I are adjacent. Then we have α(ERq) ≥ q − q + q, as desired.

4. The Unitary Polarity Graph of PG(2, q) In the next proof, we use the fact that all polarity graphs of projective planes or order q have diameter 2, no 4-cycles, and that all vertices have degree q + 1 except for the absolute points, which have degree q.

Proof of Theorem 7. Let I be an independent set of maximum size. Let EI be the edges of Uq with an endvertex in I, and degI (v) be the number of neighbors in I of a vertex v, respectively. Let a be the number of absolute points contained in I (this implies a ≤ q3/2 + 1). Then

EI = (q + 1)(|I| − a) + qa = (q + 1)|I| − a.

We wish to count the number of 2-paths which have both endpoints in I. As Uq |I| has diameter 2 and no 4-cycles, we must have exactly 2 such paths. We obtain the equation: |I| X |ΓI (v)| X deg (v) = = I 2 2 2 v∈V v∈V By Jensen’s inequality we have |I| X deg (v) EI /(|V | − |I|) = I ≥ (|V | − |I|) 2 2 2 v∈V

((q + 1)|I| − a)/(q2 + q + 1 − |I|) = (q2 + q + 1 − |I|) 2

((q + 1)|I| − q3/2 − 1)/(q2 + q + 1 − |I|) ≥ (q2 + q + 1 − |I|) 2 Equality holds throughout if and only if a = q3/2 + 1. This leads to the inequality

|I|3 + 2q|I|2 − f(q)|I| + g(q) ≤ 0. where f(q) = q3 +2q5/2 +q2 +3q3/2 +3q +3 and g(q) = (q3/2 +1)(q2 +q3/2 +q +2). 3/2 3/2 The largest root of this equation is precisely q + 1, therefore α(Uq) ≤ q + 1. 3/2 If equality holds, a = q + 1, and therefore I is the set of absolute points of Uq. INDEPENDENCE NUMBER OF PROJECTIVE NORM AND RELATED GRAPHS 11

5. The Hypergraph Hq 2 3/2 Proof of Theorem 8. We are to show that α(Hq) ≤ q /2 + q + O(q). To o facilitate the proof we will use a few facts about the graph ERq: the graph is q + 1 regular on q2 +q +1 vertices, every edge which does not have an absolute endvertex is contained in a unique triangle (see [19]), and a nonabsolute vertex is connected to either 0 or 2 absolute vertices. Recall that the vertex set of Hq is the set of nonabsolute points in V (ERq). Let I be an independent set in Hq and let J = V (Hq)\I. By Theorem 1 we have that q + 1 √ e(I,I) ≥ |I|2 − q|I|. q2 + q + 1 Similarly, q + 1 √ e(J, J) ≥ |J|2 − q|J|. q2 + q + 1 As every vertex has degree at most q + 1, we must have q + 1 √ e(I,J) ≤ |J|(q + 1) − |J|2 + q|J|. q2 + q + 1 As every edge in I lies in a unique triangle, we must have that e(I,I) ≤ e(I,J), and this leads to: q + 1 √ q + 1 √ |I|2 − q|I| ≤ |J|(q + 1) − |J|2 + q|J|. q2 + q + 1 q2 + q + 1

Substituting |J| = q2 − |I| and multiplying by q2 + q + 1 we have: √ (2q + 2) |I|2 + −q3 + 2q + 1 |I| − q2(q5/2 + q2 + q3/2 + 2q + q + 1) ≤ 0.

The roots of this equation are:

q3 − 2q − 1 ± pf(q) 4(q + 1) where

6 11 5 9 4 7 3 5 2 f(q) = q + 8q 2 + 8q + 16q 2 + 20q + 16q 2 + 22q + 8q 2 + 12q + 4q + 1.

2 3/2 It follows then from the series expansion of this root that α(Hq) ≤ q /2+q +O(q).

6. Acknowledgments The second author would like to thank Felix Lazebnik who was his thesis advisor during a portion of this research. Thanks also to Gary Ebert for fruitful discussions about Finite Geometry and norms, Chris Godsil and Mike Newman for their discussions on eigenvalue bounds, and to Benny Sudakov for bringing the problem to the attention of Felix Lazebnik, who in turn passed it on to the author. The authors would also like to thank the referees for their helpful comments. 12 DHRUV MUBAYI∗ AND JASON WILLIFORD

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(Dhruv Mubayi) Department of Mathematics, Statistics, and Computer Science, Uni- versity of Illinois, Chicago, IL 60607

(Jason Williford) Department of Mathematics and Computer Science, Albion College, Albion, MI 49224